JEE Main 202628 Jan, Morning shiftMathematicsEasyDifferential Equations163× since 2002Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation xdydx−sin2y=x3(2−x3)cos2yx\frac{dy}{dx}-\sin 2y=x^3(2-x^3)\cos^2 yxdxdy−sin2y=x3(2−x3)cos2y, x≠0x\ne0x=0. If y(2)=0y(2)=0y(2)=0, then tan(y(1))\tan(y(1))tan(y(1)) is equal toAOption A. −74-\frac{7}{4}−47BOption B. 74\frac{7}{4}47COption C. −34-\frac{3}{4}−43DOption D. 34\frac{3}{4}43CheckSkip · show solution
